Problem: Solve for $z$, $- \dfrac{6}{3z + 1} = \dfrac{1}{8} $
Solution: Multiply both sides of the equation by $3z + 1$ $ -6 = \dfrac{3z + 1}{8} $ Multiply both sides of the equation by $8$ $ -48 = 3z + 1 $ $-48 = 3z + 1$ $-49 = 3z$ $3z = -49$ $z = -\dfrac{49}{3}$